The distribution of uniformly random rotation of a i.i.d. Gaussian vector $\mathbf{x}$ given $\mathbf{x}$

Suppose that I have vector $\mathbf{x}$ that contains $n$ independently and identically distributed (i.i.d.) zero-mean Gaussian random variables $x_i\sim\mathcal{N}(0,\sigma^2)$.

Also suppose I have a uniform random rotation defined by matrix $\mathbf{R}$ that changes the angle of $\mathbf{x}$ with respect to each basis vector in the space $\mathbb{R}^n$ by some amount drawn uniformly at random from $[0,2\pi]$.

I am interested in the conditional distribution of $\mathbf{y}=\mathbf{Rx}$ given $\mathbf{x}$. It seems to me that $\mathbf{y}$ should contain i.i.d. zero-mean Gaussian random variables $y_i\sim\mathcal{N}(0,\sigma^2)$. Is that true? If so, how does one prove it?

I know that a vector of i.i.d. Gaussians is invariant to the rotation. I also know that any given rotation is a linear transformation, so if we know $\mathbf{R}$, then, obviously, $\mathbf{y}$ given $\mathbf{x}$ is deterministic and not random. I am wondering about the case when $\mathbf{R}$ is random. This question arises out of a study of a very strange interference channel in information theory. I appreciate any hints.

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If R is uniform, then, conditionally on x, y=Rx is uniformly distributed on the sphere centered at 0 with radius |x|. – Did Jul 6 '12 at 17:26
Calling this $R$ uniformly random is rather misleading. In particular, for a fixed $x$, $Rx$ is not uniformly distributed over the surface of the sphere. (Think of choosing a point on the surface of the earth; your approach is to choose the latitude and longitude uniformly. But this means that with probability $1/2$, you will be in that band that's within $\pi/4$ of the equator, and this contains more than half the earth's area.) – Nate Eldredge Jul 6 '12 at 18:58
Note that the conditional distribution of $\mathbf{y}$ given $\mathbf{x}$ cannot be normal, since $|\mathbf{y}| = |\mathbf{x}|$. – Nate Eldredge Jul 6 '12 at 18:59
(I deleted my previous comment) @did, thanks now I understand what you meant, and yes, you are correct. – M.B.M. Jul 6 '12 at 19:00
@NateEldredge When I was writing the question, I was (implicitly) using the wikipedia definition of uniform random matrices, where all the angles of the rotation w/r to the basis are uniformly random. In that setting, did's comment is correct, I think. – M.B.M. Jul 6 '12 at 19:03